Problem: Divide the following complex numbers. $ \dfrac{-20-12i}{-4-4i}$
Explanation: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-4+4i}$ $ \dfrac{-20-12i}{-4-4i} = \dfrac{-20-12i}{-4-4i} \cdot \dfrac{{-4+4i}}{{-4+4i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-20-12i) \cdot (-4+4i)} {(-4-4i) \cdot (-4+4i)} = \dfrac{(-20-12i) \cdot (-4+4i)} {(-4)^2 - (-4i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-20-12i) \cdot (-4+4i)} {(-4)^2 - (-4i)^2} = $ $ \dfrac{(-20-12i) \cdot (-4+4i)} {16 + 16} = $ $ \dfrac{(-20-12i) \cdot (-4+4i)} {32} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-20-12i}) \cdot ({-4+4i})} {32} = $ $ \dfrac{{-20} \cdot {(-4)} + {-12} \cdot {(-4) i} + {-20} \cdot {4 i} + {-12} \cdot {4 i^2}} {32} $ Evaluate each product of two numbers. $ \dfrac{80 + 48i - 80i - 48 i^2} {32} $ Finally, simplify the fraction. $ \dfrac{80 + 48i - 80i + 48} {32} = \dfrac{128 - 32i} {32} = 4-i $